# Showing that $\pi_2 (\mathbb S^2)$ is infinite using homotopy pointed class of maps $\mathbb S^2 \to \mathbb S^2$

I'd like to understand the proof of $$\pi_n(\mathbb{S}^n)\simeq \mathbb{Z}$$ if $$n \geq 1$$, (related Suspension homomorphism) The proof is "linear" but I don't understand the key part :

Thanks to Freudenthal's theorem we have that $$\pi_{n}(\mathbb{S}^n) \simeq \pi_{n+1}(\mathbb{S}^{n+1})$$ if $$n \geq 2$$ so in order to prove the thesis, is sufficient to show that $$\pi_2(\mathbb{S}^2)$$ is infinite, since again by Fredenthal $$\pi_1({\mathbb{S}^1}) \longmapsto \pi_2(\mathbb{S}^2)$$ is a surjective homomorphism.

To accomplish this, in my notes the following fact is used : "we use that we can associated to homotopy pointed classes of maps from $$\mathbb{S}^2$$ to $$\mathbb{S}^2$$, i.e $$[\mathbb{S}^2,\mathbb{S}^2]^0$$ an homomorphism in $$\text{Hom}(H_2(\mathbb{S}^2))\longmapsto \text{Hom}(H_2(\mathbb{S}^2))$$.

Apparently this homomorphism is surjective since if we think $$\mathbb{S}^2 = \mathbb{C} \cup \lbrace \infty \rbrace \longmapsto \mathbb{C} \cup \lbrace \infty \rbrace$$ sending $$z \to z^n$$ and $$\lbrace \infty \rbrace \to \lbrace \infty \rbrace$$ this induces the map $$H_2(\mathbb{S}^2) \longmapsto H_2(\mathbb{S}^2)$$ such that $$1 \to n$$.

I don't understand how to prove that the "connecting" map from $$[\mathbb{S}^2,\mathbb{S}^2]^0$$ to $$\text{Hom}(H_2(\mathbb{S}^2),H_2(\mathbb{S}^2))$$ is an homomorphism and how to see that the map sending $$z \to z^n$$ and $$\lbrace \infty \rbrace \to \lbrace \infty \rbrace$$ induces the multiplication by $$n$$.

Any help,hint or reference would be appreciated.

You don’t actually need to use that the “connecting map” $$[S^2,S^2]^0 \rightarrow \mathrm{Hom}(H_2(S^2),H_2(S^2))$$ to be a homomorphism. You just need to show that it has infinite image.
Now, we can replace $$S^2$$ with $$\mathbb{CP}^1=\mathbb{C} \cup \{\infty\}$$. Let $$f: z \longmapsto z^n$$. We can write $$\mathbb{CP}^1=U \cup V$$ with $$U=\{|z| \leq 1\}$$, $$V=\{|z|>0\} \cup \{\infty\}$$, and thus use the Mayer-Vietoris sequence for the action of $$f$$ on $$0=H_2(U\cap V) \rightarrow 0=H_2(U) \oplus H_2(V) \rightarrow H_2(U\cup V) \rightarrow H_1(U \cap V) \rightarrow 0=H_1(U) \oplus H_1(V)$$.
This works because $$U \cap V$$ is homotopic to $$S^1$$ and $$U,V$$ are contractible.
But let $$g: S^1 \rightarrow S^1$$ be given by $$z \rightarrow z^n$$, then we have an isomorphism $$H_1(S^1) \rightarrow H_1(U \cap V)$$ where $$f$$ acts on the right and $$g$$ on the left. But $$g$$ acts by multiplication by $$n$$, so that $$f$$ acts on $$H_1(U\cap V)$$ by multiplication by $$n$$. So $$f$$ acts on $$H_2(S_2) = H_2(U \cup V) \cong H_1(U \cap V)$$ by multiplication by $$n$$.